Question: $f\,^{\prime}(x)=\dfrac{24}{x^3}$ and $f(2)=12$. $f(-1) = $
Solution: Finding $f(x)$ We have $f'(x)=\dfrac{24}{x^3}$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int \dfrac{24}{x^3}\,dx \\\\ & = {-12x^{-2}} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(2)=12$. Here's what we get when we plug in $2$ : $\begin{aligned}f(2)&={-12(2)^{-2}} {+ C}\\\\ &={-3} {+ C} \end{aligned}$ We are given that this must equal $12$ : $12 = {-3} {+ C}$ Solving the equation gives us ${C=15}$. Finding $f(-1)$ Now, we have that $f(x)= {-12x^{-2}} {+15}$. Let's find $f(-1)$ by plugging in $-1$ : $\begin{aligned}f(-1)&=-12(-1)^{-2}+15\\\\ &=3 \end{aligned}$ The answer $f(-1) = 3$